**Solve the following problem :**

In the following probability distribution of a r.v.X.

x |
1 | 2 | 3 | 4 | 5 |

P (x) |
`(1)/(20)` | `(3)/(20)` | a | 2a | `(1)/(20)` |

Find a and obtain the c.d.f. of X.

#### Solution

Since the given table represents a p.m.f. of r.v. X,

\[\sum\limits_{x=1}^{5}\text{P}(x) = 1\]

∴ P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 1

∴ `(1)/(20) + (3)/(20) + "a" + 2"a" + (1)/(20)` = 1

∴ 3a = `1 - (5)/(20)`

∴ 3a = `1 - (1)/(4)`

∴ 3a = `(3)/(4)`

∴ a = `(1)/(4)`

By definition of c.d.f.,

F(x) = P(X ≤ x)

F(1) = P(X ≤ 1) = P(1) = `(1)/(20)`

F(2) = P(X ≤ 2) = F(1) + P(2)

= `(1)/(20) + (3)/(20) = (4)/(20)`

F(3) = P(X ≤ 3)

= F(2) + P(3)

= `(4)/(20) + "a" = (4)/(20) + (1)/(4) = (4)/(20) + (5)/(20) = (9)/(20)`

F(4) = P(X ≤ 4)

= F(3) + P(4)

= `(9)/(20) + 2"a" = (9)/(20) + (1)/(2) = (9)/(20) + (10)/(20) = (19)/(20)`

F(5) = P(X ≤ 5)

= F(4) + P(5)

= `(19)/(20) + (1)/(20)` = 1

∴ c.d.f. of X is as follows:

x_{i} |
1 | 2 | 3 | 4 | 5 |

F(x_{i}) |
`(1)/(20)` | `(4)/(20)` | `(9)/(20)` | `(19)/(20)` | 1 |